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Nina [5.8K]
3 years ago
14

How do I do these picture included. I will give brainliest if correct answer

Mathematics
2 answers:
natima [27]3 years ago
6 0
<span>Copy past because before, for some reason, I couldn't finish my answer.

I can only solve Q5 because I don't know all the numbers for Q9. 

1. Take +5 and do the inverse operation, which would be -5. Therefore, subtract 5 by 83. 13x + (5 - 5) = 83 - 5 ⇒ 13x = 78

<span>2. Isolate x to find x. You do this by taking 13 and dividing it by itself and 78 (We divide it because originally, you had to multiply 13 by an unknown number. The opposite operation for multiplication is division.) 

(13/13)x = 78 ⇒ (1)x [note that 1x is just like saying x] = 6. Therefore 'x' is 6. 
3. Check your work by substituting x with 6 and see if it checks out. Solve the edited problem by PEMDAS.
13(6) + 5 = 83
78 + 5 = 83
83 = 83
It checks out.
6 being x is correct.</span><span>
</span></span>
Harlamova29_29 [7]3 years ago
3 0
(5.)
13x + 5 = 83
        -5     -5
13x =78
/13   /13
x = 6

(9.)
0.7 , 0.9 , 1.6 , 2.3 , 2.5

median = 1.6
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What's the answer????
lana66690 [7]

Answer:

7

Step-by-step explanation:

2(3)(0) + 7

6(0) + 7

0 + 7

7

Have an amazing day!

<u><em>PLEASE RATE!</em></u>

8 0
2 years ago
Consider the transpose of Your matrix A, that is, the matrix whose first column is the first row of A, the second column is the
Zarrin [17]

Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.

Step-by-step explanation:

To determine if solution exist or not, you test the equation for consistency.

A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.

Consider the following scenarios:

(1) For example:Given the matrix A=\left[\begin{array}{ccc}1&2\\3&4\end{array}\right], to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:

Let A transpose be B.

∵B=\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]

the system Bx=0 can be represented in matrix form as:

\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]=\left[\begin{array}{ccc}0\\0\end{array}\right] ................................eq(1)

Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,

|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).

Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have B_{0}:      

B_{0}=\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]. The rank of B_{0} can be found by using the second column and third column pair as follows:

|B_{0}|=(3*0)-(0*2)=0 i.e, B_{0} is a singular matrix with rank of order 1.

Note: a matrix is singular if its determinant is = 0 and non-singular if it is \neq0.

Comparing the rank of both B and B_{0}, it is obvious that

Rank of B\neqRank of B_{0} since (-2)<1.

Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution.     </em>

(2) If B=\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right] is the transpose of matrix A=\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right], then

Then the equation Bx=0 is represented as:

\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]=\left[\begin{array}{ccc}0\\0\end{array}\right]..................................eq(2)

|B|= (-4*10)-(5*(-8))= -40+40 = 0  i.e B has a rank of order 1.

B_{0}=\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right],

|B_{0}|=(5*0)-(0*10)=0-0=0   i.e B_{0} has a rank of order 1.

we can therefor conclude that since

rank B=rank B_{0}=1,  equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown (X_{1} and X_{2}).

<u>Summary:</u>

  • Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
  • Determine the rank of both the coefficients matrix B and B_{0} which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
  • If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.
5 0
3 years ago
What is the average of the following numbers: 18, 64, 23, 78, 82, 91
Katen [24]
To Find The Answer, We Need To Add All Of Them Together, Then Divide By the Number Of Values. So:
64+23+78+82+91
-----------------------
             5

<span>64+23+78+82+91 = 338.
</span>
338
-----
  5

338/5 = 67.6

So, The Average Is 67.6
3 0
3 years ago
What is -4a+1a in addition
Alex

Answer:

-3a

Step-by-step explanation:

Ignore the variable since they are the same, apple and apples, and simply pay attention to the coefficient

8 0
3 years ago
Identify the type of function represented by the equation y = – 3.
snow_lady [41]

I think the answer of this question is number b. quadratic

7 0
3 years ago
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